Solve for $x$ : $6x^2 - 12x - 18 = 0$
Explanation: Dividing both sides by $6$ gives: $ x^2 {-2}x {-3} = 0 $ The coefficient on the $x$ term is $-2$ and the constant term is $-3$ , so we need to find two numbers that add up to $-2$ and multiply to $-3$ The two numbers $-3$ and $1$ satisfy both conditions: $ {-3} + {1} = {-2} $ $ {-3} \times {1} = {-3} $ $(x {-3}) (x + {1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -3) (x + 1) = 0$ $x - 3 = 0$ or $x + 1 = 0$ Thus, $x = 3$ and $x = -1$ are the solutions.